Prove that- x - 0 - lim - - 1 x - - 2 x - - 15 x - - -120-

Let L= x→ 0 ^ + lim ( [ 1 x #160; ] +[ 2 x #160; ] +......+[ 15 x #160; ] #160; ) x→ 0 ^ + lim 1 x is just greater than 1 , ...

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Byju's Answer

Standard XII

Mathematics

Evaluation of Limit

Prove that: ...

Question

Prove that:
$l i m x \to 0^{+} [[\frac{1}{x}] + [\frac{2}{x}] + . . . . + [\frac{15}{x}]] = 120.$

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Solution

Let $L = l i m x \to 0^{+} ([\frac{1}{x}] + [\frac{2}{x}] + . . . . . . + [\frac{15}{x}])$

$lim x \to 0^{+} \frac{1}{x}$ is just greater than $1$ , $lim x \to 0^{+} \frac{2}{x}$ is just greater than $2$ and so on

$L = 1 + 2 + 3 + . . . . . . + 15$

$∴ L = 120$

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Prove that:limx→0+[[1x]+[2x]+....+[15x]]=120.

Let L=limx→0+([1x]+[2x]+......+[15x])limx→0+1x is just greater than 1, limx→0+2x is just greater than 2 and so onL=1+2+3+......+15∴L=120

Prove that:
$l i m x \to 0^{+} [[\frac{1}{x}] + [\frac{2}{x}] + . . . . + [\frac{15}{x}]] = 120.$

Let $L = l i m x \to 0^{+} ([\frac{1}{x}] + [\frac{2}{x}] + . . . . . . + [\frac{15}{x}])$

$lim x \to 0^{+} \frac{1}{x}$ is just greater than $1$ , $lim x \to 0^{+} \frac{2}{x}$ is just greater than $2$ and so on

$L = 1 + 2 + 3 + . . . . . . + 15$

$∴ L = 120$